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Germain primes

  1. Mar 30, 2008 #1
    Reference: www.mathpages.com/home367.htm
    On page 2 of reference the formula is given
    (x+y)^p - x^p - y^p = pxy(x=y)Q(x,y) where Q(x,y) is a homogenous integer function of degree p-3.
    If we insert a number of different value of p into the equation, it appears that
    Q(x,y) = (x^2 = xy + y^2)^((p-3)/2)

    Is there an easy way to prove this without getting lost in infinite series calculations, or is there a proof already in print?
  2. jcsd
  3. Mar 30, 2008 #2
    You seem in a much better position to investigate this than anybody else, particularly since your reference can not be found.

    (x+y)^p - x^p - y^p = pxy(x=y)Q(x,y) where Q(x,y) is a homogenous integer function of degree p-3.

    Are you sure you mean to write x=y? If so what is the point of Q(x,y)? Assuming you don't mean x=y, the Q(x,y) appears to be of degree p-2. Because we subtracted those of degree p, and pulled out a pxy from Q(x,y,) that leaves degree p-2.
    Last edited: Mar 30, 2008
  4. Mar 31, 2008 #3
    Sorry for the errors. The correct web address is

    The equal sign in the formula should have been "+" not "=". I neglected to hit the shift key.
    I think p-3 is correct. When you substact x^p and y^p from the expansion, the results have a factor of pxy, so the x^p in now at the p-2 level. However, the remaining equation is divisible by (x+y), and this brings it to the p-3 level.

    I'm an amateur looking for professional help. Is there an easy way to prove that

    Q(x,y) = (x^2 + xy + y^2)^((n-3)/2) ?
  5. Mar 31, 2008 #4
    I think you are assuming too much from other people here. It does help to state the definition of Q(x,y)

    [tex]Q(x,y) = \frac{(x+y)^P-(x^P+y^P)}{(xy)(x+y)p}[/tex]

    I thought it enough of a problem, letting p-1 = u, to show that p divides all terms: [tex](x+y)^u -\frac{x^u+y^u}{x+y}[/tex]
    This can be found by induction on k: [tex]\frac{(p-1)!}{k!(p-1-k)!} \equiv (-1)^k Mod p [/tex]

    So you maybe looking at an induction problem on n.
    Last edited: Mar 31, 2008
  6. Mar 31, 2008 #5
    Thank you. It's very much appreciated.
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